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3 years ago

Is .09 greater then .0003

Mathematics

1 answer:

Vlada [557]3 years ago

8 0

Answer:

Yes

Step-by-step explanation:

For ease, let's multiply these by 10,000.

0.09 • 10,000 = 900.00 = 900

0.0003 • 10,000 = 3.00 = 3

900 is clearly greater than 3. Therefore, 0.09 is greater than 0.0003.

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Prove that sinxtanx=1/cosx - cosx

maks197457 [2]

Answer:

See below

Step-by-step explanation:

We want to prove that

$\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}$

Taking the RHS, note

$\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}$

Remember that

$\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)$

Therefore,

$\dfrac{1-\cos^2(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)} = \dfrac{\sin(x)\sin(x)}{\cos(x)}$

Once

$\dfrac{\sin(x)}{\cos(x)} = \tan(x)$

Then,

$\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)$

Hence, it is proved

5 0

2 years ago

Write a quadratic function in standard form with zeros 1 and -10

Licemer1 [7]

Answer:

f(x)=x^2+9x-10

Step-by-step explanation:

<u>Standard Form of Quadratic Function</u>

The standard form of a quadratic function is:

$f(x)=ax^2+bx+c$

where a,b, and c are constants.

The factored form of a quadratic equation is:

$f(x)=a(x-\alpha)(x-\beta)$

Where $\alpha$ and $\beta$ are the roots or zeros of f, and a is constant.

We know the zeros of the function are 1 and -10. The function is:

$f(x)=a(x-1)(x-(-10))$

$f(x)=a(x-1)(x+10)$

Operating:

$f(x)=a(x^2+10x-x-10)$

Joining like terms:

$f(x)=a(x^2+9x-10)$

Since we are not given any more restrictions, we can choose the value of a=1, thus. the required function is:

$\boxed{f(x)=x^2+9x-10}$

6 0

3 years ago

Alon was curious if segments \overline{JK} JK start overline, J, K, end overline and \overline{NM} NM start overline, N, M, end

BartSMP [9]

The answer is A because Alon could have rotated the figure and discovered that they are congruent

6 0

3 years ago

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

e-lub [12.9K]

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$

5 0

3 years ago

Ellen is playing a video game in which she captures butterflies. There are 6 butterflies onscreen, but the number of butterflies

Anna11 [10]

Answer:

(2 x 4) - 8

Step-by-step explanation:

4 0

2 years ago